A parabola has a focus of (4,2) and a directrix of x=-2. Which equation can represent the parabola?

Question

A. (x+2)^2 = 6(y+1)

B. (y-2)^2 = 24(x-1)

C. (y-2)^2 = 12(x-1)

D. (x+2)^2 =3(y+1)

Answer 1

C is correct

focus is 6 to the right of the directrix line, so the parabola opens to the right.

General form of that type of parabola is x-h=a(y-k)² (h,k) is the vertex. Vertex is half way or the midpont

between the focus and the directrix or (1,2) x-1=a(y-2)²

The vertex (1,2) is a point on the parabola. Try substituting that into any of A, B, C, or D and only B or C might work. Only B and C are even of the correct general form. (1,2) does satisfy C. A and D are the worst answers,as they are up and down parabolas, opening up and down.

(4,2) is the focus. The parabola is all points (x,y) equidistant from the directrix line x=-2 and the focus.

Distance from (x,y) to (4,2) is the hypotenuse of a right triangle with sides y-2 and x-4 Sum their squares then take the sum's square root to get the hypotenuse. Set it equal to x+2, the distance to the directrix.

Then (x+2)² = (y-2)² + (x-4)² or

x² +4x +4 = y² -4y +4 + x² -8x +16

12x-12 = y² - 4 y + 4 = (y-2)²

12(x-1) = (y-2)² That's answer C

Let y= 6, then x= 7/3 (7/3,6) is a point on the parabola. It is 13/3 from the directrix line and it's distance to the focus (4,2) is the square root of the sum of sides 4-7/3=5/3 and 6-2=4, each side squared

(25/9+ 144/9)^1/2 = 1/3. That's a check on the answer, using a point on the parabola.

Answer 2

The correct answer is C. Based on the location of the focus and directrix, the vertex must be at (1,2) and it must open to the right. The distance between the vertex and the focus is 3 and the center to the directrix is also 3. This means p=3 and 4p=12. All this is contained in the following equation.

C. (y-2)² = 12(x-1)